Difference between revisions of "2002 AIME II Problems/Problem 8"
I like pie (talk | contribs) m |
Jackshi2006 (talk | contribs) (→Solution 3) |
||
(10 intermediate revisions by 9 users not shown) | |||
Line 1: | Line 1: | ||
− | |||
== Problem == | == Problem == | ||
+ | Find the least positive integer <math>k</math> for which the equation <math>\left\lfloor\frac{2002}{n}\right\rfloor=k</math> has no integer solutions for <math>n</math>. (The notation <math>\lfloor x\rfloor</math> means the greatest integer less than or equal to <math>x</math>.) | ||
== Solution == | == Solution == | ||
− | {{solution}} | + | ===Solution 1=== |
+ | Note that if <math>\frac{2002}n - \frac{2002}{n+1}\leq 1</math>, then either <math>\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor</math>, | ||
+ | or <math>\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor+1</math>. Either way, we won't skip any natural numbers. | ||
+ | |||
+ | The greatest <math>n</math> such that <math>\frac{2002}n - \frac{2002}{n+1} > 1</math> is <math>n=44</math>. (The inequality simplifies to <math>n(n+1)<2002</math>, which is easy to solve by trial, as the solution is obviously <math>\simeq \sqrt{2002}</math>.) Note Once we plug in a few values we can see easily that it is <math>\boxed{049}.</math> | ||
+ | |||
+ | |||
+ | We can now compute: | ||
+ | <cmath>\left\lfloor\frac{2002}{45}\right\rfloor=44 </cmath> | ||
+ | <cmath>\left\lfloor\frac{2002}{44}\right\rfloor=45 </cmath> | ||
+ | <cmath>\left\lfloor\frac{2002}{43}\right\rfloor=46 </cmath> | ||
+ | <cmath>\left\lfloor\frac{2002}{42}\right\rfloor=47 </cmath> | ||
+ | <cmath>\left\lfloor\frac{2002}{41}\right\rfloor=48 </cmath> | ||
+ | <cmath>\left\lfloor\frac{2002}{40}\right\rfloor=50 </cmath> | ||
+ | |||
+ | From the observation above (and the fact that <math>\left\lfloor\frac{2002}{2002}\right\rfloor=1</math>) we know that all integers between <math>1</math> and <math>44</math> will be achieved for some values of <math>n</math>. Similarly, for <math>n<40</math> we obviously have <math>\left\lfloor\frac{2002}{n}\right\rfloor > 50</math>. | ||
+ | |||
+ | Hence the least positive integer <math>k</math> for which the equation <math>\left\lfloor\frac{2002}{n}\right\rfloor=k</math> has no integer solutions for <math>n</math> is <math>\boxed{049}</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | Rewriting the given information and simplifying it a bit, we have | ||
+ | <cmath> \begin{align*} | ||
+ | k \le \frac{2002}{n} < k+1 &\implies \frac{1}{k} \ge \frac{n}{2002} > \frac{1}{k+1}. \\ &\implies \frac{2002}{k} \ge n > \frac{2002}{k+1}. | ||
+ | \end{align*} </cmath> | ||
+ | |||
+ | Now note that in order for there to be no integer solutions to <math>n,</math> we must have <math>\left\lfloor \frac{2002}{k} \right\rfloor = \left\lfloor \frac{2002}{k+1} \right\rfloor.</math> We seek the smallest such <math>k.</math> A bit of experimentation yields that <math>k=49</math> is the smallest solution, as for <math>k=49,</math> it is true that <math>\left\lfloor \frac{2002}{49} \right\rfloor = \left\lfloor \frac{2002}{50} \right\rfloor = 40.</math> Furthermore, <math>k=49</math> is the smallest such case. (If unsure, we could check if the result holds for <math>k=48,</math> and as it turns out, it doesn't.) Therefore, the answer is <math>\boxed{049}.</math> | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | In this solution we use inductive reasoning and a lot of trial and error. Depending on how accurately you can estimate, the solution will come quicker or slower. | ||
+ | |||
+ | Using values of k as 1, 2, 3, 4, and 5, we can find the corresponding values of n relatively easily. For k = 1, n is in the range [2002-1002]; for k = 2, n is the the range [1001-668], etc: 3, [667,501]; 4, [500-401]; 5, [400-334]. For any positive integer k, n is in a range of <math>floor[2002/k]-ceiling[2002/(k+1)]</math>. | ||
+ | |||
+ | Now we try testing k = 1002 to get a better understanding of what our solution will look like. Obviously, there will be no solution for n, but we are more interested in how the range will compute to. Using the formula we got above, the range will be 1-2. Testing any integer k from 1002-2000 will result in the same range. Also, notice that each and every one of them have no solution for n. Testing 1001 gives a range of 2-2, and 2002 gives 1-1. They each have a solution for n, and their range is only one value. Therefore, we can assume with relative safety that the integer k we want is the lowest integer that follows this equation: | ||
+ | |||
+ | floor[2002/k] + 1 = ceiling[2002/(k+1)] | ||
+ | |||
+ | Now we can easily guess and check starting from k = 1. After a few tests it's not difficult to estimate a few jumps, and it took me only a few minutes to realize the answer was somewhere in the forties. Then it's just a matter of checking them until we get <math>\boxed{049}</math>. | ||
+ | Alternatively, you could use the equation above and proceed with one of the other two solutions listed. | ||
+ | |||
+ | |||
+ | -jackshi2006 | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=II|num-b=7|num-a=9}} | {{AIME box|year=2002|n=II|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:46, 12 December 2020
Problem
Find the least positive integer for which the equation has no integer solutions for . (The notation means the greatest integer less than or equal to .)
Solution
Solution 1
Note that if , then either , or . Either way, we won't skip any natural numbers.
The greatest such that is . (The inequality simplifies to , which is easy to solve by trial, as the solution is obviously .) Note Once we plug in a few values we can see easily that it is
We can now compute:
From the observation above (and the fact that ) we know that all integers between and will be achieved for some values of . Similarly, for we obviously have .
Hence the least positive integer for which the equation has no integer solutions for is .
Solution 2
Rewriting the given information and simplifying it a bit, we have
Now note that in order for there to be no integer solutions to we must have We seek the smallest such A bit of experimentation yields that is the smallest solution, as for it is true that Furthermore, is the smallest such case. (If unsure, we could check if the result holds for and as it turns out, it doesn't.) Therefore, the answer is
Solution 3
In this solution we use inductive reasoning and a lot of trial and error. Depending on how accurately you can estimate, the solution will come quicker or slower.
Using values of k as 1, 2, 3, 4, and 5, we can find the corresponding values of n relatively easily. For k = 1, n is in the range [2002-1002]; for k = 2, n is the the range [1001-668], etc: 3, [667,501]; 4, [500-401]; 5, [400-334]. For any positive integer k, n is in a range of .
Now we try testing k = 1002 to get a better understanding of what our solution will look like. Obviously, there will be no solution for n, but we are more interested in how the range will compute to. Using the formula we got above, the range will be 1-2. Testing any integer k from 1002-2000 will result in the same range. Also, notice that each and every one of them have no solution for n. Testing 1001 gives a range of 2-2, and 2002 gives 1-1. They each have a solution for n, and their range is only one value. Therefore, we can assume with relative safety that the integer k we want is the lowest integer that follows this equation:
floor[2002/k] + 1 = ceiling[2002/(k+1)]
Now we can easily guess and check starting from k = 1. After a few tests it's not difficult to estimate a few jumps, and it took me only a few minutes to realize the answer was somewhere in the forties. Then it's just a matter of checking them until we get . Alternatively, you could use the equation above and proceed with one of the other two solutions listed.
-jackshi2006
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.